List Beautifier

My solution:

def listBeautifier(a):
    res = a[:]
    while res and res[0] != res[-1]:
       b, *res, c = res
    return res

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Mex Function

My solution:

def mexFunction(s, upperBound):
    found = -1
    for i in range(upperBound):
        if not i in s:
            found = i
            break
    else:
        found = upperBound

    return found

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Base Conversion

My solution:

def baseConversion(n, x):
    return hex(sum((int(n[i]) if n[i] <= '9' else ord(n[i]) - ord('a') + 10) * x ** (len(n) - 1 - i) for i in range(len(n))))[2:]

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Simple Sort

My solution:

def simpleSort(arr):

    n = len(arr)

    for i in range(n):
        j = 0
        stop = n - i
        while j < stop - 1:
            if arr[j] > arr[j + 1]:
                arr[j + 1], arr[j] = arr[j], arr[j + 1]
            j += 1
    return arr

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Modulus

My solution:

def modulus(n):
    if type(n) == int:
        return n % 2
    else:
        return -1

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Count Bits

My solution:

def countBits(n):
    return n.bit_length()

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Language Differences

My answer: "x = 15, y = -4"

Output:
Python 3:

-4
3
1
0
-4

Java:

-4
3
1
0
-3

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Special Conditional

My answer: "a == not b"
Two operators ("=="  and "not") cannot be next to each other.

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Efficient Comparison

My answer: "Option 1 is the most optimal one."

My solution:

import time
L = 1
x = 2
y = 3
R = 9
t1 = time.perf_counter()
def a():
 if L < x ** y <= R:
  return True

t2 = time.perf_counter()
print(t2 - t1)
t3 = time.perf_counter()
def b():
 if x ** y > L and x ** y <= R:
  return True

t4 = time.perf_counter()
print(t4 - t3)
t5 = time.perf_counter()
def c():
 if x ** y in range(L + 1, R + 1):
  return True

t6 = time.perf_counter()
print(t6 - t5)

Output:

5.131850057605017e-07
1.5395550172814964e-06
1.5395550172815045e-05

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Collections Truthness

My answer: [True, False]
Explanation on 4.1 Truth Value Testing.

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